Measure theory Study Guide

📖 Core Concepts Measure – A set function \(\mu:\mathcal{A}\to[0,\infty]\) assigning a “size” to each measurable set, generalizing length, area, volume, mass, or probability. Measurable Space – Pair \((X,\mathcal{A})\) where \(X\) is a set and \(\mathcal{A}\) is a σ‑algebra (closed under complements and countable unions). Measure Space – Triple \((X,\mathcal{A},\mu)\) adding a measure \(\mu\) to a measurable space. Null Set – A measurable set \(N\) with \(\mu(N)=0\). Complete Measure – Every subset of a null set is also measurable. Probability Measure – A measure with total mass \(\mu(X)=1\). Signed / Complex Measure – Allow negative or complex values; signed drops non‑negativity, complex requires finite total variation. σ‑finite Measure – \(X=\bigcup{k=1}^\infty Ek\) with each \(\mu(Ek)<\infty\). --- 📌 Must Remember Non‑negativity: \(\mu(A)\ge 0\) for all \(A\in\mathcal{A}\). Null empty set: \(\mu(\varnothing)=0\). Countable Additivity: For disjoint \(\{Ai\}\), \[ \mu\Bigl(\bigcup{i=1}^\infty Ai\Bigr)=\sum{i=1}^\infty \mu(Ai). \] Monotonicity: \(A\subseteq B \implies \mu(A)\le\mu(B)\). Countable Subadditivity: Always have \[ \mu\Bigl(\bigcup{i=1}^\infty Ai\Bigr)\le\sum{i=1}^\infty \mu(Ai). \] Continuity from Below: If \(A1\subseteq A2\subseteq\cdots\), then \[ \mu\Bigl(\bigcup{n=1}^\infty An\Bigr)=\lim{n\to\infty}\mu(An). \] Continuity from Above: If \(A1\supseteq A2\supseteq\cdots\) and some \(Ak\) has finite measure, then \[ \mu\Bigl(\bigcap{n=1}^\infty An\Bigr)=\lim{n\to\infty}\mu(An). \] σ‑finite ≠ Finite: A σ‑finite measure can be infinite (e.g., Lebesgue measure on \(\mathbb{R}\)). Completeness: Adding all subsets of null sets to \(\mathcal{A}\) yields the completion of the measure. --- 🔄 Key Processes Verify a Set Function is a Measure Check the domain is a σ‑algebra. Confirm non‑negativity and \(\mu(\varnothing)=0\). Prove countable additivity for any disjoint collection. Show Continuity from Below Establish an increasing sequence \(An\). Use monotonicity to get \(\mu(An)\) increasing and bounded above by \(\mu(\cup An)\). Apply the definition of limit to conclude equality. Show Continuity from Above Ensure a decreasing sequence \(An\) with at least one finite \(\mu(Ak)\). Apply the complement trick: \(\bigcap An = X\setminus\bigcup (X\setminus An)\). Use continuity from below on the complements. Complete a Measure Identify all null sets \(N\). Extend \(\mathcal{A}\) to \(\mathcal{A}^\ = \{A\cup S: A\in\mathcal{A},\ S\subseteq N,\ \mu(N)=0\}\). Define \(\mu^\(A\cup S)=\mu(A)\). --- 🔍 Key Comparisons Measure vs. Signed Measure Measure: \(\mu(A)\ge0\) for all \(A\). Signed Measure: may take negative values; still countably additive. Probability Measure vs. General Measure Probability: \(\mu(X)=1\). General: \(\mu(X)\) can be any non‑negative (possibly infinite) number. σ‑finite vs. Finite Finite: \(\mu(X)<\infty\). σ‑finite: \(X\) = countable union of finite‑measure pieces (allows \(\mu(X)=\infty\)). Complete Measure vs. Incomplete Measure Complete: every subset of a null set is measurable. Incomplete: some subsets of null sets are not in \(\mathcal{A}\). Content (Finitely Additive) vs. Measure (Countably Additive) Content: additive only for finite disjoint unions. Measure: additive for countable disjoint unions. --- ⚠️ Common Misunderstandings “Additivity holds for overlapping sets.” Additivity (countable) requires pairwise disjoint sets; overlapping sets need inclusion‑exclusion. “Every subset of a null set is automatically measurable.” Only true after completion; the original σ‑algebra may omit such subsets. “σ‑finite means the total measure is finite.” σ‑finite allows infinite total measure; the key is a countable cover by finite‑measure pieces. “Countable subadditivity is an equality.” It is an inequality; equality needs disjointness (additivity). --- 🧠 Mental Models / Intuition Measure ≈ “Generalized length.” Imagine stretching a rubber sheet over a set; the amount of rubber needed is its measure. Null sets are “dust.” They’re so thin that they can be ignored for “almost everywhere” statements. σ‑finite = “Finite on each tile of a floor.” The whole floor may be infinite, but you can tile it with countably many finite‑size pieces. Completion = “Adding the hidden crumbs.” You fill in the missing subsets of dust so nothing is left out. --- 🚩 Exceptions & Edge Cases Non‑measurable Sets: Vitali set, Banach–Tarski pieces—exist by the Axiom of Choice; cannot be assigned a Lebesgue measure. Continuity from Above fails if no \(Ak\) has finite measure (e.g., decreasing sets each of infinite measure). Countable Subadditivity can be strict: \(\mu\bigl(\bigcup Ai\bigr) < \sum \mu(Ai)\) when overlaps are present. Signed/Complex Measures may not be positive; the Jordan decomposition separates a signed measure into its positive and negative parts. --- 📍 When to Use Which Probability problems → use a probability measure (total mass = 1). Radon–Nikodym theorem → require σ‑finite measures on both sides. Spectral theorem → employ projection‑valued (operator‑valued) measures. When only finite additivity is needed (e.g., elementary length assignments) → a content suffices. Handling differences of measures → switch to a signed measure (or Jordan decomposition). --- 👀 Patterns to Recognize Increasing sequence \(\Rightarrow\) apply continuity from below. Decreasing sequence with a finite‑measure term \(\Rightarrow\) apply continuity from above. Zero‑measure sets appearing in statements → think “almost everywhere”; the conclusion can ignore them. Problem mentions “cover by countably many sets of finite measure” → the underlying measure is σ‑finite. If a question asks whether a function is integrable w.r.t. a measure → check that the set where the function is infinite is a null set. --- 🗂️ Exam Traps Distractor: “\(\mu(A\cup B)=\mu(A)+\mu(B)\) for any sets \(A,B\).” Why tempting: Mirrors additivity, but forgets the disjointness requirement. Distractor: “Every subset of a null set is measurable by definition.” Why tempting: Confuses complete measures with arbitrary measures. Distractor: “If \(\mu(X)=\infty\) the measure cannot be σ‑finite.” Why tempting: Misinterprets σ‑finite as “finite overall”. Distractor: “A content is a measure because it is additive.” Why tempting: Overlooks the missing countable additivity. Distractor: “Non‑measurable sets cannot exist if the axiom of choice is rejected.” Why tempting: The axiom of choice creates such sets; without it they may not be provable, but the statement is too absolute. ---